External (6 credits)

1. Mechanics 2.4 External (6 credits)

2. Scalars and Vectors • Vector quantities have magnitude (size) and direction e.g. displacement, velocity. • Scalar quantities only have magnitude, e.g. distance, speed.

3. Displacement

4. Displacement • Displacement is a vector quantity. It is the distance and direction from a known origin (starting point). Displacement is measured in m. e.g. if you travel 4 km North, then 3 km West, your displacement (from your starting position) is 5 km at 37° West of North. 3 km 5 km 4 km 37°

5. Scalars and Vectors This process of adding two (or more) vector quantities is called vector addition. The vectors must be of the same quantity e.g. both displacement vectors or both velocity vectors Vectors are added head-to-tail.

6. Vector Addition watch how the ORDER of vector addition does not matter!

7. HMWK pg 37-38

8. Displacement-Time Graphs A displacement-time graph has displacement on the y-axis and time on the x-axis. Displacement (m) Time (s)

9. Displacement-Time Graphs Here is a displacement-time graph of a person who walks forward 3 m in a time of 3 s, stands still for 3 s, then turns around and takes 6 s to walk the 3 m back to their starting point. Displacement (m) Time (s)

10. Displacement-Time Graphs  The gradient of a displacement time graph gives the ________of the object. velocity Displacement (m) Time (s)

12. Velocity

13. Starter Apollo 11, the first spacecraft to land people on the Moon, took 102 hours to reach its destination. The average speed of the craft was 5500 km/h. What is the distance to the Moon? distance = speed x time = 5500 km x 102 hr hr = 561 000 km

14. Velocity Velocity is a vector quantity. It is the speed and the direction from a known origin. Velocity is calculated by the change in displacement divided by the time taken. v = Dd Dt v = velocity, m.s-1 Dd = change in displacement, m Dt = change in time, s

15. Motion of an object can be described using the vector quantities: • displacement • distance and direction from a known origin • velocity • the speed and direction in which an object moves • acceleration • the change in velocity per unit time a.= Δv Δt

16. e.g. 8 km A car travels 6 km north in 4 minutes and then 8 km east in 5 minutes. Find their displacement, average velocity and their average speed. 6 km 10 km N 53° tanθ = Δd = √62 + 82 Displacement is 10 km at a bearing of 053° = √100 θ = tan-1 = 10 θ = 53.13° Average Velocity Average Speed vavg. = Δd Δt .= Δdistance Δtime .= 14 km 0.15 hr Δt = 9 min = 10/.15 = 0.15 hr = 93.3 km/hr = 66.7 km/hr

17. Recall: Displacement-Time Graph  The gradient of a displacement time graph gives the ________of the object. velocity Displacement (m) Time (s)

18. Velocity -Time Graphs  A velocity time graph has velocity on the y-axis and time on the x-axis velocity (m/s) Time (s)

19. We can draw one using information from the other Displacement (m) Time (s) velocity (m/s) Time (s)

20. Motion GraphsVelocity - Time  The gradient of a velocity - time graph gives the ___________ of the object. acceleration Velocity (m/s) Time (s)   Astraight linerepresents constant acceleration

21. Motion GraphsVelocity - Time a. = Δv Δt   The acceleration is equal to the change in velocity divided by the change in time a. = 7− 2 10 7 Velocity (m/s) a. = 0.5 m/s2 2 Time (s) 10   Astraight linerepresents constant acceleration

22. Motion GraphsVelocity - Time  The area under a velocity - time graph gives the ________________of the object. total displacement Velocity (m/s) Time (s)   Astraight linerepresents constant acceleration

23. Motion GraphsVelocity - Time Distance travelled = (2 × 14) + ½ (14 × 5) = 63 m 7 Velocity (m/s) 2 Time (s) 14 initial velocity

24. Motion GraphsVelocity - Time  The gradient of a velocity - time graph gives the ___________ of the object.  The area under a velocity - time graph gives the ______________ . acceleration distance travelled   Astraight linerepresents constant acceleration

25. Acceleration Acceleration is a vector quantity. It is the change in velocity per unit time. It is measured in Acceleration is calculated by the change in velocity divided by the change in time. a = Dv Dt a = acceleration, m.s-2 Dv = change in velocity, m.s-1 Dt = change in time, s

26. Acceleration Acceleration due to gravity near the surface of the earth is 9.81 Acceleration due to gravity near the surface of the moon is 1.63

27. Starter A car starts moving from rest at a stop sign and accelerates at 1.5 m/s2. a.) Calculate the speed of the bus after 7 seconds. b.) After how many seconds is the bus travelling at a speed of 18 m/s? c.) If the car is traveling at 20m/s and comes to a stop in 5 s, what is the braking deceleration? a = Dv Dt v = a × t v = 1.5 m/s2× 7 s = 10.5 m/s t= v a t= 18 1.5 t = 12 s a = 0 - 20 5 a = - 4 m/s2

28. Motion GraphsVelocity - Time Draw a velocity time graph for a car initially moving at a constant velocity of 20 m/s for 3 seconds and then decelerates at a rate of 2.5 m/s2. At what time is the car at rest? Calculate the total displacement of the car from your graph. 3 x 20 + ½ (11-3) x 20 = 60 + 80 = 140 m Velocity (m/s) 30 20 10 Time (s) 3 11

29. Kinematics The geometry of motion

30. Kinematics Equations The kinematics equationsare a series of equations linking displacement, initial velocity, final velocity, time taken, and acceleration. These equations only hold in situations where the acceleration is constant e.g. constant acceleration due to gravity. The kinematics equations can be derived from the equations for velocity and acceleration.

31. Kinematics Equations vf = vi + at d = ½ (vi + vf)t d = vit + ½ at2 vf2 = vi2+ 2ad d = displacement, m vi (or u) = initial velocity, m.s-1 vf (or v) = final velocity, m.s-1 a = acceleration, m.s-2 t = time, s.

32. Example A motorcycle accelerates from rest to 25m/s in 22 seconds. a.) Calculate the acceleration. b.) Calculate the distance travelled. c.) The bike brakes to a stop in 5 seconds. Calculate is deceleration. d.) Calculate the distance travelled while braking. vf- vi = a t vf = vi + at a = 1.14 m/s2 vf = vi + at d = 275 m d = ½ (vi + vf)t vf- vi = a t 0 - 25= a 5 vf = vi + at a = -5 m/s2 d = vit + ½ at2 d = 62.5 m d =25(5) + ½ (-5)(52)

33. Gravity g = 9.81 ms-2 = 9.81 Nkg-1 ‘ each kg feels a weight force of 9.81 N’

34. Weight m = mass (kg) g = -9.81 ms-2 W = mg * we often assume the direction ‘up’ is positive the length of time it takes an object to fall is independent of it mass (assuming there is no air resistance)

35. Example vi = 0 m/s This boy drops a ball off a 80 m cliff. How long will it take to reach the bottom? d = vit + ½ at2 -80 = 0 + ½ (-9.81)t2 -80x2 = t2 -9.81 What is the balls speed when it reaches the bottom? 16.31 = t2 t = 4.04 s vf = vi + at = 0– 9.81 x 4.04 vf = - 39.6 m/s

36. Example vi = 0 m/s How far has the ball fallen after only 2.5 seconds? d = vit + ½ at2 = 0 + ½ (-9.81)(2.5)2 = - 30.7 m How long does it take to reach a speed of -25 m/s? vf = vi + at -25 = 0 – 9.81 t t = -25_ -9.81 t = 2.55 s

37. pg 32-33 Example A 1.5kg firecracker is fired vertically upwards with an initial speed of 40ms-1. Calculate the time to reach maximum height. vf = vi + at 0= 40 – 9.81 t t = -40_ -9.81 t = 4.08 s What is its acceleration at the top of its flight? g = -9.81 m/s2 Calculate the weight force, W, on the firecracker just before it lands. W = -14.7 N W = mg W = 1.5 x -9.81

38. Projectile Motion A projectile is an object that is travelling through the air. The object experiences the force of gravity, but we assume there are no other forces acting on it (i.e. we assume there is no friction). We can consider the horizontal and vertical components of projectile motion separately.

39. Projectile Motion Since there are no horizontal forces acting on the object, the horizontal component of the velocity remains constant. The horizontal displacement can be calculated using v = Dd Dt The vertical component of the velocity changes at the acceleration due to gravity, so the vertical displacement and velocity can be calculated using the kinematics equations.

40. Projectile Motion Since there are no horizontal forces acting on the object, the horizontal component of the velocity remains constant. The horizontal displacement can be calculated using v = Dd Dt The vertical component of the velocity changes at the acceleration due to gravity, so the vertical displacement and velocity can be calculated using the kinematics equations. The resultant displacement and resultant velocity can be calculated by vector addition of the horizontal and vertical components.

41. Average Speed The average speed of an object is the total distancetravelled divided by the time taken. Both average speed and total distance are scalar quantities. vavg= d t vavg = average speed, m.s-1 d = total distance, m t = time taken, s.

42. Relative Motion Relative velocity is the velocity of an object relative to another.

43. Relative Motion The red car, travelling at 80 km/hr, approaches the green car travelling at 50 km/hr. Find the velocity of the red car relative to the green car. vgreen relative red = vgreen −vred 50 km.hr = 50 – 80 = -30 The green car is travelling 30 km/hr backwards relative to the red car. 80 km.hr

44. 6 m.s-1 • A 8 m.s-1 Relative Motion e.g. A boat is travelling across a river with a velocity of 6 ms-1 relative to the water. The river is flowing downstream at a velocity of 8 m.s-1 relative to the banks. The velocity of the boat relative to a stationary observer on the bank can be calculated using vector addition. The resultant velocity is calculated using Pythagoras: vbA = √(vbr2 + vrA2) vbA = 10 tan q = vrA/vbr = 8/6 q = 37° The boat is moving with a velocity of 10m/s of at a bearing of 107°